Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.Use the given values of n and p to find the mean μ and standard deviation σ.Use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.n=1415, p=3/5 μ

Answer: [tex]\mu=849[/tex][tex]\sigma=18.43[/tex]Minimum usual value[tex]=\mu-2\sigma=849-2(18.43)=849-36.86=812.14[/tex]Maximum usual value[tex]=\mu+2\sigma=849+2(18.43)=849+36.86=885.86[/tex]Step-by-step explanation:The Normal approximation can be used to binomial distribution, when n is large and p is near to 0.5.[tex]X\sim N(np,\sqrt{np(1-p)})[/tex] here, Mean = [tex]\mu=np[/tex]Standard deviation :[tex]\sigma=\sqrt{np(1-p)}[/tex]Given : [tex]n= 1415,\ p=\dfrac{3}{5}[/tex]Then, [tex]\mu=1415\times\dfrac{3}{5}=849[/tex][tex]\sigma=\sqrt{1415(\dfrac{3}{5})(1-\dfrac{3}{5})}\\\\=\sqrt{339.6}\approx18.43[/tex] Range rule of thumb : It says that the range is approximately four times the standard deviation. Minimum usual value[tex]=\mu-2\sigma=849-2(18.43)=849-36.86=812.14[/tex]Maximum usual value[tex]=\mu+2\sigma=849+2(18.43)=849+36.86=885.86[/tex]